Dirichlet's test

Proof

Let $S_n=\sum _{k=1}^n{a}_k{b}_k$ and $B_n=\sum _{k=1}^n{b}_k$.

From summation by parts, we have that $S_n=a_n{B}_n+\sum _{k=1}^{n-1}{B}_k(a_k-a_{k+1})$. Since the magnitudes of the partial sums ${\displaystyle B_n}$ are bounded by some M and ${\displaystyle a_n\to 0}$ as ${\displaystyle n\to \mathrm{\infty }}$, the first of these terms approaches zero: ${\displaystyle |a_n{B}_n|\le |a_{n}M|\to 0}$ as ${\displaystyle n\to \mathrm{\infty }}$.

Furthermore, for each k, ${\displaystyle |B_k(a_k-a_{k+1})|\le M|a_k-a_{k+1}|}$.

Since ${\displaystyle (a_n)}$ is monotone, it is either decreasing or increasing:

• If ${\displaystyle (a_n)}$ is decreasing, $${\displaystyle \sum _{k=1}^{n}M|a_k-a_{k+1}|=\sum _{k=1}^{n}M(a_k-a_{k+1})=M\sum _{k=1}^n(a_k-a_{k+1}),}$$ which is a telescoping sum that equals ${\displaystyle M(a_1-a_{n+1})}$ and therefore approaches ${\displaystyle M{a}_1}$ as ${\displaystyle n\to \mathrm{\infty }}$. Thus, $\sum _{k=1}^{\mathrm{\infty }}M|a_k-a_{k+1}|$ converges.
• If ${\displaystyle (a_n)}$ is increasing, $${\displaystyle \sum _{k=1}^{n}M|a_k-a_{k+1}|=-\sum _{k=1}^{n}M(a_k-a_{k+1})=-M\sum _{k=1}^n(a_k-a_{k+1}),}$$ which is again a telescoping sum that equals ${\displaystyle -M(a_1-a_{n+1})}$ and therefore approaches ${\displaystyle -M{a}_1}$ as ${\displaystyle n\to \mathrm{\infty }}$. Thus, again, $\sum _{k=1}^{\mathrm{\infty }}M|a_k-a_{k+1}|$ converges.

So, the series $\sum _{k=1}^{\mathrm{\infty }}{B}_k(a_k-a_{k+1})$ converges by the direct comparison test to $\sum _{k=1}^{\mathrm{\infty }}M|a_k-a_{k+1}|$. Hence ${\displaystyle S_n}$ converges.

Applications

A particular case of Dirichlet's test is the more commonly used alternating series test for the case $${\displaystyle b_n=(-1{)}^n⟹\left|\sum _{n=1}^N{b}_n\right|\le 1.}$$

Another corollary is that $\sum _{n=1}^{\mathrm{\infty }}{a}_n\sin n$ converges whenever ${\displaystyle (a_n)}$ is a decreasing sequence that tends to zero. To see that ${\displaystyle \sum _{n=1}^N\sin n}$ is bounded, we can use the summation formula $${\displaystyle \sum _{n=1}^N\sin n=\sum _{n=1}^N\frac{e^{in}-e^{-in}}{2i}=\frac{\sum _{n=1}^N(e^i{)}^n-\sum _{n=1}^N(e^{-i}{)}^n}{2i}=\frac{\sin 1+\sin N-\sin (N+1)}{2-2\cos 1}.}$$

Improper integrals

An analogous statement for convergence of improper integrals is proven using integration by parts. If the integral of a function f is uniformly bounded over all intervals, and g is a bounded non-negative monotonically decreasing function, then the integral of fg is a convergent improper integral.